The radiation by an extended linear source in the unbounded uniaxial dielectric medium with uniform current distribution is studied. The exact as well as approximate fields in the near and far zones are found analytically using the dyadic Green functions in the frequency domain. Elegant closed-form results are obtained when the Hertzian dipole is parallel and perpendicular to the optic axis of the uniaxial medium. When the dipole is parallel to the optic axis, only extraordinary waves are emitted. When the dipole is perpendicular to the optic axis, both ordinary and extraordinary waves are emitted; however, the radiations are suppressed along the optic axis and no extraordinary waves are emitted in a direction perpendicular to both the electric dipole and the optic axis. A comparison with the point dipole showed that the directivity of the radiation pattern can be controlled using the length of the Hertzian dipole.
1. Introduction
A medium which has a single axis of symmetry and can support electromagnetic waves with two orthogonal states of linear polarization states having different phase velocities is called a uniaxial medium [1–3]. These mediums exist naturally, e. g., rutile and calcite [4], and they can also be fabricated artificially, e. g., a stack of thin films with alternating high and low refractive indices [5]. Uniaxial mediums are very attractive class of mediums from the point of view of applications in nano imaging [6] , nano sensing [7], nano resonators [8], thermal emission engineering [9], efficient absorbers [10], and subdiffraction imaging [11].
A small linear electric current element with uniform current distribution is called a Hertzian dipole. Several electrically small sources of electromagnetic waves can be modeled as Hertzian dipoles such as quantum dots or quantum wires. The length of the Hertzian dipole is usually short as compared to the wavelength of electromagnetic waves emitted by that dipole [12,13]. The Hertzian dipole is also highly useful for helping analyze larger antennas which can be subdivided into short sections having uniform currents as large antennas can be thought of as composed of small sections with uniform current distribution on each small section [12].
To the zeroth order, small sources of radiations can be modeled as point dipoles. Therefore, several authors have studied the radiations from point dipoles. For example, the radiation by a point dipole in unbounded uniaxial medium is dealt in Refs. [2,14,15]. The radiation over a layered medium with its optic axis lying perpendicular to the plane of stratification has been studied by Tsang et al. [16], Kong [17], Kwon and Wang [18], and Tang [19], while the same problem with point dipole embedded in the stratified medium has been studied by Ali and Mahmoud [20]. Far field radiation emitted by an arbitrarily oriented point dipole which is placed in a two-layered uniaxially anisotropic medium with its tilted optic axis is treated analytically with the use of dyadic Green function[13]. Here two cases are discussed, when the dipole is placed over the two layered uniaxial medium, and when it is embedded in a two layered uniaxial medium [13].
Most of the analytical work on radiation by sources inside the uniaxial mediums has been done for point dipoles, which are the zeroth order approximation of real sources. For the next approximation, it is important to consider the radiation from sources with finite length. The simplest case for this type of source is a Hertzian dipole with constant current density phasor over the length of the source. This is the problem addressed in this paper. However, unlike the point dipole, when the integration is trivial, the results cannot be obtained in closed form for general Hertzian dipole. Therefore, we solved for two cases of the orientation of the Hertzian dipole: (i) When the dipole is parallel to the optic axis and (ii) when the dipole is perpendicular to the optic axis. A generally oriented dipole can always be broken into two vector components, one along and one perpendicular to the optic axis. Therefore, our formulation can be used to construct the solution of generally oriented Hertzian dipole by simple vector superposition. Let us note that the extended sources have been dealt with for radiation in the uniaxial medium numerically [15] and using integral equations [21], and using approximation [22]. However, our aim is to find closed-form and rigorous solutions of an extended linear dipole in this paper.
The plan of the paper is as follow: The dyadic Green functions of the uniaxial dielectric mediums are reproduced along with their near and far-field approximations in Sec. 2. The case of the electric dipole oriented parallel to the optic axis, when both the optic axis and the dipole are parallel to the $x$ axis, is presented in Sec. 3. The case of the dipole oriented parallel to the optic axis, when both the optic axis and the dipole are parallel to the $z$ axis, is presented in Sec. 4. The case of the dipole perpendicular to the optic axis is presented in Sec. 5. The representative numerical results are discussed in Sec. 6 and are compared with a point dipole in Sec. 7. Finally, the concluding remarks are presented in Sec. 8. In this paper, $\exp (-i\omega t)$ time-dependence is implicit, where $t$ is the time, $\omega$ is the angular frequency, and $i=\sqrt {-1}$. Moreover, $\varepsilon _o$, $\mu _o$, and $k_o=\omega \sqrt {\varepsilon _o\mu _o}$, represent the free-space permittivity, permeability, and wavenumber respectively. Boldface letters represent vectors and symbols underlined twice represent dyadics. The identity dyadic is represented as $\underline {\underline {I}}$.
2. Dyadic Green function
The dyadic Green functions relate the vector source with the vector field [23] and are solution of the Maxwell equations. The dyadic Green functions for the uniaxial mediums have been derived earlier [1,2,14,23] and are reproduced here for completeness. Let us begin with the time-harmonic Maxwell equations [1,2]
(1)$$\nabla \times \mathbf{H}(\mathbf{r})+i\omega \mathbf{D}(\mathbf{r})=\mathbf{J} _{e}(\mathbf{r}),$$
(2)$$\nabla \times \mathbf{E}(\mathbf{r})-i\omega \mathbf{B}(\mathbf{r})=0,$$
(3)$$\nabla\cdot \mathbf{D}(\mathbf{r}) = \rho_{e}(\mathbf{r}),$$
(4)$$\nabla\cdot\mathbf{B}(\mathbf{r})=0,$$
where $\mathbf {J}_{e}(\mathbf {r})$ is the electric current density and $\rho _{e}(\mathbf {r})$ is the electric charge density related by the continuity equation $\nabla \cdot \mathbf{J}_e(\mathbf{r})-i\omega \rho _e(\mathbf{r})=0$. The frequency domain constitutive relations can be written as (5)$$\mathbf{D}(\mathbf{r})=\underline{\underline{\varepsilon}}\cdot{\mathbf{E}}({\mathbf{r}})\,,\quad \mathbf{B}(\mathbf{r})=\mu_{o}\mu_{b}\mathbf{H}(\mathbf{r}),$$
with the permittivity dyadic $\underline {\underline {\varepsilon }}$ given by (6)$$\underline{\underline{\varepsilon}}=\varepsilon_{o}\underline{\underline{\varepsilon}}_{r}=\varepsilon_{o}[\varepsilon_b\underline{\underline{I}}+(\varepsilon_a - \varepsilon_b)\hat{\mathbf{c}}\hat{\mathbf{c}}],$$
where $\hat {\mathbf {c}}$ is the unit vector representing the direction of the optic axis.Since Eqs. (1) and (2) are linear in $\mathbf {E}$ and $\mathbf {H}$, the electric and magnetic field phasors can be written in terms of the dyadic Green functions as
(7)$$\mathbf{E}(\mathbf{r})=\mathop{\iiint}_{V'}\underline{\underline{G}}^{ee}(\mathbf{R})\cdot\mathbf{J}_{e}(\mathbf{r'})d^{3}\mathbf{r'},$$
(8)$$\mathbf{H}(\mathbf{r})=\mathop{\iiint}_{V'}\underline{\underline{G}}^{me}(\mathbf{R})\cdot\mathbf{J}_{e}(\mathbf{r'})d^{3}\mathbf{r'},$$
where $V'$ is the volume occupied by the electric current density $\mathbf {J}_{e}(\mathbf {r'})$ and (9)$$\mathbf{R}=\mathbf{r}-\mathbf{r'}\,.$$
The electric dyadic Green function $\underline {\underline {G}}^{ee}(\mathbf {R})$ and magnetoelectric dyadic Green function $\underline {\underline {G}}^{me}(\mathbf {R})$ can be found as [1,23] (10)$$\begin{aligned} \underline{\underline{G}}^{ee}(\mathbf{R})&={i}\omega\mu_{o}\mu_{b} \Bigg\{{g}_{e}(\mathbf{R})\varepsilon_{a}\underline{\underline{\varepsilon}}_{r}^{-1}\Big[1-\frac{1}{{i}{k}_{o}{n}_{o}{R}_{e}}-\frac{1}{({k}_{o}{n}_{o}{R}_{e})^{2}}\Big] \\ &-{g}_{e}(\mathbf{R})\Big[1-\frac{3}{{i}{k}_{o}{n}_{o}{R}_{e}}-\frac{3}{({k}_{o}{n}_{o}{R}_{e})^{2}}\Big]\frac{\varepsilon_{a}^{2}(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R})(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R})}{{R}_{e}^{2}} \\ &+\frac{1}{\varepsilon_{b}}\Big[\varepsilon_b{g}_{o}(\mathbf{R})-\varepsilon_a{g}_{e}(\mathbf{R})\Big]\underline{\underline{K}}(\mathbf{R})\\ &+\frac{{R}{g}_{o}(\mathbf{R})-{R}_{e}{g}_{e}(\mathbf{R})}{i{k}_{o}{n}_{o}(\mathbf{R}\times\hat{\mathbf{c}})\cdot(\mathbf{R}\times\hat{\mathbf{c}})}\Big[\underline{\underline{I}}-\hat{\mathbf{c}}\hat{\mathbf{c}}-2\underline{\underline{K}}(\mathbf{R})\Big] \Bigg\}\,, \end{aligned}$$
(11)$$\begin{aligned} \underline{\underline{G}}^{me}(\mathbf{R})&=\frac{\varepsilon_a}{\varepsilon_b}(1-{{i}{k}_{o}{n}_{o}{R}_{e}}){g}_{e}(\mathbf{R})\frac{(\mathbf{R}\times\hat{\mathbf{c}})[\mathbf{R}\times(\mathbf{R}\times\hat{\mathbf{c}})]}{\mathit{R}_{e}^{2}(\mathbf{R}\times\hat{\mathbf{c}})\cdot(\mathbf{R}\times\hat{\mathbf{c}})}\\ &+\Big[{g}_{e}(\mathbf{R})-{g}_{o}(\mathbf{R})\Big](\mathbf{R}\cdot\hat{\mathbf{c}})\frac{[\hat{\mathbf{c}}\times(\mathbf{R}\times\hat{\mathbf{c}})](\mathbf{R}\times\hat{\mathbf{c}})+(\mathbf{R}\times\hat{\mathbf{c}})[\hat{\mathbf{c}}\times(\mathbf{R}\times\hat{\mathbf{c}})]}{[(\mathbf{R}\times\hat{\mathbf{c}})\cdot(\mathbf{R}\times\hat{\mathbf{c}})]^{2}}\\ &-(1-{{i}{k}_{o}{n}_{o}{R}}){g}_{o}(\mathbf{R})\frac{[\mathbf{R}\times(\mathbf{R}\times\hat{\mathbf{c}})](\mathbf{R}\times\hat{\mathbf{c}})}{\mathit{R}^{2}[(\mathbf{R}\times\hat{\mathbf{c}})\cdot(\mathbf{R}\times\hat{\mathbf{c}})]}\,, \end{aligned}$$
where (12)$$\underline{\underline{K}}(\mathbf{R})=\frac{(\mathbf{R}\times\hat{\mathbf{c}})(\mathbf{R}\times\hat{\mathbf{c}})}{(\mathbf{R}\times\hat{\mathbf{c}})\cdot(\mathbf{R}\times\hat{\mathbf{c}})},$$
(13)$$\underline{\underline{\varepsilon}}_{r}^{-1}=\frac{1}{\varepsilon_{b}}\underline{\underline{I}}-\Big(\frac{1}{\varepsilon_{b}}-\frac{1}{\varepsilon_{a}}\Big)\hat{\mathbf{c}}\hat{\mathbf{c}},$$
and (14)$$n_{o}=\sqrt{\varepsilon_{b}}\sqrt{\mu_{b}}, \quad k_{o}=\omega\sqrt{\mu_{o}\varepsilon_{o}}\,.$$
The scalar Green functions (15)$${g}_{o}(\mathbf{R})=\frac{\exp({i}{k}_{o}{n}_{o}{R})}{4\pi{R}} \quad \textrm{and}\quad {g}_{e}(\mathbf{R})=\frac{\exp({i}{k}_{o}{n}_{o}{R}_{e})}{4\pi{R}_{e}}\,$$
represent ordinary and extraordinary waves, respectively, and (16)$${R}_{e}=\sqrt{\frac{\varepsilon_a}{\varepsilon_b}(\mathbf{R}\times\hat{\mathbf{c}})\cdot(\mathbf{R}\times\hat{\mathbf{c}})+(\mathbf{R}\cdot\hat{\mathbf{c}})^{2}}\,.$$
Let us note that the ordinary waves propagate with the same phase velocity in all directions in the uniaxial medium and the extraordinary waves have their phase velocity dependent upon the direction of propagation which is different from that of the phase velocity of the ordinary waves unless the waves are propagating along the optic axis of the medium [1].The dyadic Green function can be approximated in the near field by retaining terms proportional to ${1}/{R^{3}}$ since we have $k_{o}n_{o}R_{e}\ll 1$ and $k_{o}n_{o}R\ll 1$, which means that we are observing radiation at a distance which is much smaller than the wavelength. Therefore, Eqs. (10) and (11) can be written as
(17)$$\underline{\underline{G}}^{ee}(\mathbf{R})\approx\frac{i\varepsilon _{a}}{4\pi \omega \varepsilon _{o}\varepsilon _{b}}\left( \frac{3\varepsilon _{a}\left( \underline{\underline{\varepsilon }}_{r}^{-1}\cdot \mathbf{R}\right) \left( \underline{\underline{\varepsilon }}_{r}^{-1}\cdot \mathbf{R}\right) }{ R_{e}^{5}}-\frac{\underline{\underline{\varepsilon }}_{r}^{-1}}{R_{e}^{3}} \right)$$
and (18)$$\underline{\underline{G}}^{me}(\mathbf{R})\approx 0\,$$
in the near field.In the far field, our point of observation is at a distance which is much greater than the wavelength of the radiation, that is, $\mathit {k}_{o}\mathit {n}_{o}\mathit {R}_{e}\gg 1$ and $\mathit {k}_{o}\mathit {n}_{o}\mathit {R}\gg 1$. Therefore, to compute the electromagnetic field in the far field, we retain the terms proportional to ${1}/{R}$. So, Eq. (10) can be approximated as
(19)$$\begin{aligned} \underline{\underline{G}}^{ee}(\mathbf{R})\approx &{i}\omega\mu_{o}\mu_{b} \Bigg\{{g}_{e}(\mathbf{R})\varepsilon_{a}\underline{\underline{\varepsilon}}_{r}^{-1}-{g}_{e}(\mathbf{R})\frac{\varepsilon_{a}^{2}(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R})(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R})}{{R}_{e}^{2}} \\ &+\frac{1}{\varepsilon_{b}}\Big[\varepsilon_b{g}_{o}(\mathbf{R})-\varepsilon_a{g}_{e}(\mathbf{R})\Big]\underline{\underline{K}}(\mathbf{R})\Bigg\}\,. \end{aligned}$$
Using the identity [1,23] (20)$$\underline{\underline{\varepsilon}}_{r}^{-1}-\frac{\varepsilon_{a}(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R})(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R})}{{R}_{e}^{2}}-\frac{1}{\varepsilon_{b}}\underline{\underline{K}}(\mathbf{R})=\frac{[\mathbf{R}\times(\mathbf{R}\times\hat{\mathbf{c}})][\mathbf{R}\times(\mathbf{R}\times\hat{\mathbf{c}})]}{\varepsilon_{b} R_{e}^{2}(\mathbf{R}\times\hat{\mathbf{c}})\cdot(\mathbf{R}\times\hat{\mathbf{c}})},$$
Eq. (19) can be rearranged as (21)$$\underline{\underline{G}}^{ee}(\mathbf{R})=i\omega\mu_{o}\mu_{b}\Bigg\{g_{e}(\mathbf{R})\frac{\varepsilon_{a}[\mathbf{R}\times(\mathbf{R}\times\hat{\mathbf{c}})][\mathbf{R}\times(\mathbf{R}\times\hat{\mathbf{c}})]}{\varepsilon_{b} R_{e}^{2}(\mathbf{R}\times\hat{\mathbf{c}})\cdot(\mathbf{R}\times\hat{\mathbf{c}})}+g_{o}(\mathbf{R})\underline{\underline{K}}(\mathbf{R})\Bigg\}\,.$$
Similarly, Eq. (11) can be written as (22)$$\underline{\underline{G}}^{me}(\mathbf{R})\approx{i}{k}_{o}{n}_{o}\Bigg\{{g}_{o}(\mathbf{R})\frac{[\mathbf{R}\times(\mathbf{R}\times\hat{\mathbf{c}})](\mathbf{R}\times\hat{\mathbf{c}})}{\mathit{R}(\mathbf{R}\times\hat{\mathbf{c}})\cdot (\mathbf{R}\times\hat{\mathbf{c}})}-\frac{\varepsilon_a}{\varepsilon_b}{g}_{e}(\mathbf{R})\frac{(\mathbf{R}\times\hat{\mathbf{c}})[\mathbf{R}\times(\mathbf{R}\times\hat{\mathbf{c}})]}{\mathit{R}_{e}[(\mathbf{R}\times\hat{\mathbf{c}})\cdot(\mathbf{R}\times\hat{\mathbf{c}})]}\Bigg\}\,$$
in the far field.3. Dipole and optic axis parallel to x axis
Let us begin with the simpler case when both the electric dipole and the optic axis are parallel to the $x$ axis as shown in Fig. 1 with the electric current density
(23)$$\mathbf{J}_{e}(\mathbf{r'})= \begin{cases} -i\omega p_{o}\delta(y')\delta(z')\hat{\mathbf{x}}\,, & \quad |x'|\leq L\,,\\ \quad 0\,, & \quad |x'|>L\,. \end{cases}$$
3.1 Near field
The near-field of the Hertzian dipole parallel to the optic axis can be computed by substituting the near-field approximation of the dyadic Green function (17) and the electric current density (23) into Eq. (7) to get
(24)$$\mathbf{E}(\mathbf{r})=\frac{\varepsilon_{d}{p}_{o}}{4\pi\varepsilon_{o}}\int_{-L}^{L}\Bigg[\frac{3\varepsilon_{a}(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R}_{x})(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R}_{x})\cdot\hat{\mathbf{x}}}{{R}_{ex}^{5}}-\frac{\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\hat{\mathbf{x}}}{{R}_{ex}^{3}}\Bigg]dx^{\prime },$$
where (25)$$\varepsilon_{d}=\varepsilon_{a}/\varepsilon_{b},$$
(26)$${R}_{ex}=\sqrt{\varepsilon _{d}(y^{2}+z^{2})+(x-x')^{2}},$$
and (27)$$\mathbf{R_{x}}=(x-x')\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}\,.$$
Using Eqs. (13) and (27), we can get the following identities (28)$$\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R}_{x}=\frac{1}{\varepsilon_{b}}\left({y}\hat{\mathbf{y}}+{z}\hat{\mathbf{z}}\right)+\frac{1}{\varepsilon_{a}} ({x}-{x}^{'})\hat{\mathbf{x}},$$
(29)$$\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\hat{\mathbf{x}}=\frac{1}{\varepsilon_{a}}\hat{\mathbf{x}}\,.$$
Substituting Eqs. (28) and (29) into Eq. (24), we get (30)$$\begin{aligned} \mathbf{E}(\mathbf{r})=&\frac{\varepsilon_{d}p_{o}}{4\pi\varepsilon_{o}\varepsilon_{a}}\Bigg\{3\varepsilon_{d}(y\hat{\mathbf{y}}+z\hat{\mathbf{z}})\int_{-L}^{L}\frac{({x}-{x}^{'})}{\left[\varepsilon_{d}(y^{2}+z^{2})+(x-x^{\prime })^{2}\right]^{\frac{5}{2}}}dx^{'}\\ &+3\hat{\mathbf{x}}\int_{-L}^{L}\frac{({x}-{x}^{'})^{2}}{\left[\varepsilon_{d}(y^{2}+z^{2})+(x-x^{'})^{2}\right]^{\frac{5}{2}}}dx^{'}-\hat{\mathbf{x}}\int_{-L}^{L}\frac{1}{\left[\varepsilon_{d}(y^{2}+z^{2})+(x-x^{'})^{2}\right]^{\frac{3}{2}}}dx^{'}\Bigg\}\, \end{aligned}$$
that gives (31)$$\begin{aligned} \mathbf{E}(\mathbf{r})=&\frac{\varepsilon_{d}p_{o}}{4\pi\varepsilon_{o}\varepsilon_{b}}\Bigg\{y\hat{\mathbf{y}}+z\hat{\mathbf{z}}\Bigg(\frac{1}{\left[\varepsilon_{d} (y^{2}+z^{2})+(x-L)^{2}\right]^{\frac{3}{2}}}-\frac{1}{\left[\varepsilon_{d} (y^{2}+z^{2})+(x+L)^{2}\right]^{\frac{3}{2}}}\Bigg)\\ &+\frac{\hat{\mathbf{x}}}{\varepsilon_{d}}\Bigg(\frac{L-x}{\left[\varepsilon_{d} (y^{2}+z^{2})+(x-L)^{2}\right]^{\frac{3}{2}}}+\frac{L+x}{\left[\varepsilon_{d} (y^{2}+z^{2})+(x+L)^{2}\right]^{\frac{3}{2}}}\Bigg)\Bigg\}\,. \end{aligned}$$
Since $\mathbf {H}(\mathbf {r})=0$ in the near-field, Eq. (31) effectively represents the electrostatic field of a line of charge in the uniaxial medium parallel to the optic axis.3.2 Far field
The electric and magnetic fields in the far field can be found by the substitution of Eqs. (19) and (23) into Eq. (7), to get
(32)$$\mathbf{E}(\mathbf{r})=\omega^{2}\mu_{o}\mu_{b}{p}_{o}\Bigg\{\int_{-L}^{L}{g}_{e}(\mathbf{R}_{x})\Bigg[\varepsilon_{a}\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\hat{\mathbf{x}}-\frac{\varepsilon_{a}^{2}(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R}_{x})(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R}_{x})\cdot\hat{\mathbf{x}}}{{R}_{ex}^{2}}\Bigg]dx^{\prime}\Bigg\},$$
which can be written as (33)$$\begin{aligned} \mathbf{E}(\mathbf{r})=&\frac{\omega^{2}\mu_{o}\mu_{b}p_{o}}{4\pi}\Bigg[\hat{\mathbf{x}}\int_{-L}^{L}\frac{\exp({i}{k}_{o}{n}_{o}{R}_{ex})}{{R}_{ex}}dx^{\prime}-\varepsilon_{d}(y\hat{\mathbf{y}}+z\hat{\mathbf{z}})\int_{-L}^{L}\frac{(x-x^{\prime})\exp({i}{k}_{o}{n}_{o}{R}_{ex})}{{R}_{ex}^{3}}dx^{\prime}\\ &-\hat{\mathbf{x}}\int_{-L}^{L}\frac{(x-x^{\prime})^{2}\exp({i}{k}_{o}{n}_{o}{R}_{ex})}{{R}_{ex}^{3}}dx^{\prime}\Bigg]\,, \end{aligned}$$
using Eqs. (15), (28) and (29), where $R_{ex}$ is given by Eq. (26).Since our point of observation is far away from the dipole, i.e., $r_{e}\gg x^{\prime }$, which means that our distance from the dipole is much greater than the size of the dipole. So, in this limit, we neglect the square and higher order terms in the binomial expansion of Eq. (26) and approximate as
(34)$${R}_{ex}\simeq{r}_{e}-\frac{x}{{r}_{e}}x'\,$$
in the exponential factor, where (35)$${r}_{e}=\sqrt{x^{2}+\varepsilon_{d}(y^{2}+z^{2})}\,.$$
In the denominator of Eq. (34), we can approximate ${R}_{ex}\simeq {r}_{e}$. Thereafter, Eq. (33) becomes (36)$$\mathbf{E}(\mathbf{r})\simeq\frac{\omega^{2}\mu_{o}\mu_{b}p_{o}}{4\pi}\exp({i}{k}_{o}{n}_{o}{r}_{e})\Bigg[\left(\frac{I_{1}}{r_{e}}-\frac{I_{3}}{r_{e}^{3}}\right)\hat{\mathbf{x}}-\varepsilon_{d}\Big(\frac{y\hat{\mathbf{y}}+z\hat{\mathbf{z}}}{{r}_{e}^{3}}\Big)I_{2}\Bigg],$$
where the integrals $I_{1}$ to $I_{3}$ are derived in the appendix.Using the transformation from Cartesian to spherical coordinates as
(37)$$x=r\sin\theta\cos\phi ,\quad y=r\sin\theta\sin\phi, \quad z=r\cos\theta,$$
we can rewrite Eq. (35) as (38)$$r_{e}=r\sqrt{\sin^{2}\theta\cos^{2}\phi+\varepsilon_{d}\Big(\sin^{2}\theta\sin^{2}\phi+\cos^{2}\theta\Big)}=r\Phi(\theta,\phi)\,.$$
Now by substituting the solutions of integrals $I_{1}$, $I_{2}$ and $I_{3}$ along with Eqs. (37) and (38) into Eq. (36), the final expression of the electric field in spherical coordinates is given by (39)$$\begin{aligned} \mathbf{E}(\mathbf{r})=&\frac{k_{o}\mu_{b}p_{o}\varepsilon_{d}}{2\pi\varepsilon_{o}n_{o}r}\frac{\exp(ik_{o}n_{o}r\Phi)}{\Phi^2}\Bigg[\frac{(\sin^2\theta\sin^2\phi+\cos^2\theta)}{\sin\theta\cos\phi}\hat{\mathbf{x}}\\ &-\sin\theta\sin\phi\hat{\mathbf{y}}-\cos\theta\hat{\mathbf{z}}\Bigg]\sin\Big(\frac{k_{o}L n_{o}\sin\theta\cos\phi}{\Phi}\Big)\,. \end{aligned}$$
The expression of the magnetic field can be found by using Eqs. (22) and (23) in Eq. (8) as (40)$$\mathbf{H}(\mathbf{r})=-{k}_{o}{n}_{o}\omega{p}_{o}\varepsilon_{d}\int_{-L}^{L}{g}_{e}(\mathbf{R}_{x})\frac{(\mathbf{R}_{x}\times\hat{\mathbf{x}})[\mathbf{R}_{x}\times(\mathbf{R}_{x}\times\hat{\mathbf{x}})]\cdot\hat{\mathbf{x}}}{\mathit{R}_{ex}(\mathbf{R}_{x}\times\hat{\mathbf{x}})\cdot(\mathbf{R}_{x}\times\hat{\mathbf{x}})}dx^{\prime}\,.$$
Using Eq. (15) we get (41)$$\mathbf{H}(\mathbf{r})=\frac{{k}_{o}{n}_{o}\omega{p}_{o}\varepsilon_{d}}{4\pi}(-y\hat{\mathbf{z}}+z\hat{\mathbf{y}})\int_{-L}^{L}\frac{\exp({i}{k}_{o}{n}_{o}{R}_{ex})}{\mathit{R}_{ex}^{2}}dx^{\prime},$$
after making use of the following (42)$$\mathbf{R}_{x}\times\mathbf{\hat{\mathbf{x}}}=-y\hat{\mathbf{z}}+z\hat{\mathbf{y}}, \quad [\mathbf{R}_{x}\times(\mathbf{R}_{x}\times\mathbf{\hat{\mathbf{x}}})]\cdot\hat{\mathbf{x}}=-(y^{2}+z^{2})\,.$$
Using Eqs. (34) in the exponential term and $R_{ex}=r_e$ in the denominator of Eq. (41), we get (43)$$\mathbf{H}(\mathbf{r})=\frac{{k}_{o}{n}_{o}\omega{p}_{o}\varepsilon_{d}}{4\pi}\frac{\exp({i}{k}_{o}{n}_{o}{r}_{e})}{{r}_{e}^{2}}(-y\hat{\mathbf{z}}+z\hat{\mathbf{y}})I_{1}\,.$$
By substituting the value of $I_{1}$ into Eq. (43), transforming to the spherical coordinates and ignoring $1/r^{2}$ terms we get (44)$$\mathbf{H}(\mathbf{r})=\frac{k_{o}p_{o}c\varepsilon_{d}}{2\pi r}\frac{\exp(ik_{o}n_{o}r\Phi)}{\Phi\sin\theta\cos\phi}\Big(\cos\theta\hat{\mathbf{y}}-\sin\theta\sin\phi\hat{\mathbf{z}}\Big)\sin\Big(\frac{k_{o}L n_{o}\sin\theta\cos\phi}{\Phi}\Big)\,.$$
From Eqs. (39) and (44), we can see that only extraordinary waves propagate in this case. Furthermore, the electric and magnetic fields are perpendicular to each other, as is usually the case of radiation in the far zone. The time-averaged power radiated per unit solid angle by the Hertzian dipole is given by [24,25] (45)$$\frac{dP}{d\Omega}=\frac{1}{2}\mathbf{\hat{r}}\cdot\rm{Re}(\mathbf{E}\times\mathbf{H^{*}})r^{2}\,.$$
Using the expressions for $\mathbf {E}$ and $\mathbf {H}$ from Eqs. (39) and (44) in Eq. (45), we get (46)$$\frac{dP}{d\Omega}=\frac{k_{o}^{2}\mu_{b}p_{o}^{2}c\varepsilon_{d}^2}{8\pi^{2}\varepsilon_{o}n_{o}}\frac{(\sin^2\theta\sin^2\phi+\cos^2\theta)}{\Phi^3\sin^2\theta\cos^2\phi}\sin^2\Bigg(\frac{k_{o}L n_{o}\sin\theta\cos\phi}{\Phi}\Bigg)\,.$$
When we substitute $\varepsilon _{d}=1$, $\Phi =1$, we get the results for the isotropic medium.4. Dipole and optic axis parallel to z axis
In the previous section, we had the dipole and the optic axis parallel to the $x$ axis. In the following section, we will present the results for the dipole to be parallel to the $z$ axis, but the optic axis would still be parallel to the $x$ axis. The reason is that an arbitrarily oriented dipole can be resolved into a dipole parallel to and perpendicular to the optic axis. Therefore, the results of the previous section and the following section can be used to construct solutions of an arbitrarily oriented Hertzian dipole. However, in this section, we present the results of for a Hertzian dipole when the optic axis is parallel to the $z$ axis so that the results of the next section and this section can be compared for the case of isotropic medium for consistency check.
When the dipole and the optic axis are parallel to the $z$ axis as shown in Fig. 2, the expression for the electric field is given by
(47)$$\mathbf{E}(\mathbf{r})=\frac{k_{o}\mu_{b}p_{o}\varepsilon_{d}}{2\pi\varepsilon_{o}n_{o}r}\frac{\exp(ik_{o}n_{o}r\Theta)}{\Theta^2}\Bigg[\frac{\sin^2\theta}{\cos\theta}\mathbf{\hat{x}}-\sin\theta\cos\phi\mathbf{\hat{y}}-\sin\theta\sin\phi\mathbf{\hat{z}}\Bigg]\sin\Bigg(\frac{k_{o}L n_{o}\cos\theta}{\Theta}\Bigg),$$
where $\Theta$ is (48)$$\Theta=\sqrt{\cos^2\theta+\varepsilon_{d}\sin^2\theta},$$
and the expression for magnetic field is given by (49)$$\mathbf{H}(\mathbf{r})=\frac{k_{o}p_{o}c\varepsilon_{d}}{2\pi r}\frac{\exp(ik_{o}n_{o}r\Theta)}{\Theta\cos\theta}\Big(-\sin\theta\cos\phi\mathbf{\hat{z}}+\sin\theta\sin\phi\mathbf{\hat{y}}\Big)\sin\Bigg(\frac{k_{o}L n_{o}\cos\theta}{\Theta}\Bigg)\,.$$
The power distribution then is given by (50)$$\frac{dP}{d\Omega}=\frac{k_{o}^{2}\mu_{b}p_{o}^{2}c\varepsilon_{d}^2}{8\pi^{2}\varepsilon_{o}n_{o}}\frac{\sin^2\theta}{\Theta^3\cos^2\theta}\sin^2\Bigg(\frac{k_{o}L n_{o}\cos\theta}{\Theta}\Bigg)\,.$$
The result is independent of $\phi$ as should be the case since the dipole and the optic axis are both along the $z$ axis and the problem has azimuthal symmetry. The far field for the isotropic medium can be obtained by setting $\varepsilon _{d}=1$ and $\Theta =1$.5. Dipole parallel to z axis and optic axis parallel to x axis
Let us now consider the case when the Hertzian dipole is perpendicular to the optics axis, as shown schematically in Fig. 3. For this electric dipole, the current density can be written as
(51)$$\mathbf{J}_{e}(\mathbf{r})= \begin{cases} -i\omega p_{o}\delta(x')\delta(y')\hat{\mathbf{z'}}, & |z'|\leq L\,,\\ \quad 0\,, & |z'|> L\,. \end{cases}$$
5.1 Near field
In the near field, the expression of electric field is found by substitution of Eqs. (17) and (51) into Eq. (7) as
(52)$$\mathbf{E}(\mathbf{r})=\frac{\varepsilon_{d}p_{o}}{4\pi\varepsilon_{o}}\int_{-L}^{L}\Bigg[\frac{3\varepsilon_{a}(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R}_{z})(\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\mathbf{R}_{z})\cdot\hat{\mathbf{z}}}{{R}_{ez}^{5}}-\frac{\underline{\underline{\varepsilon}}_{r}^{-1}\cdot\hat{\mathbf{z}}}{{R}_{ez}^{3}}\Bigg]dz^{\prime },$$
where (53)$$\mathbf{R}_{z}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+(z-z')\hat{\mathbf{z}}\,$$
and (54)$$R_{ez}=\sqrt{x^{2}+\varepsilon_{d}y^{2}+\varepsilon_{d}(z-z')^{2}}\,.$$
After simplification, we get $$\begin{aligned}\mathbf{E}(\mathbf{r})=&\frac{3\varepsilon_{d}^{2}{p}_{o}}{4\pi\varepsilon_{o}\varepsilon_{a}}\Bigg\{\Big(x\hat{\mathbf{x}}+\varepsilon_{d}y\hat{\mathbf{y}}\Big)\int\limits_{-L}^{L}\frac{(z-z^{\prime})}{[x^2+\varepsilon_{d} y^2+\varepsilon_{d}(z-z^{\prime})^{2}]^\frac{5}{2}}dz^{\prime}\\&+\hat{\mathbf{z}}\varepsilon_{d}\int\limits_{-L}^{L}\frac{(z-z^{\prime})^{2}}{[x^2+\varepsilon_{d} y^2+\varepsilon_{d}(z-z^{\prime})^{2}]^\frac{5}{2}}dz^{\prime}-\frac{\hat{\mathbf{z}}}{3}\int\limits_{-L}^{L}\frac{1}{[x^2+\varepsilon_{d} y^2+\varepsilon_{d}(z-z^{\prime})^{2}]^\frac{3}{2}}dz^{\prime}\Bigg\},\end{aligned}$$
(55)$$\begin{aligned} \mathbf{E}(\mathbf{r})=&\frac{3\varepsilon_{d}^{2}{p}_{o}}{4\pi\varepsilon_{o}\varepsilon_{a}}\Bigg[\Big(x\hat{\mathbf{x}}+\varepsilon_{d}y\hat{\mathbf{y}}\Big)\Bigg\{\frac{1}{[x^2+\varepsilon_{d} y^2+\varepsilon_{d}(z-L)^{2}]^\frac{3}{2}}-\frac{1}{[x^2+\varepsilon_{d} y^2+\varepsilon_{d}(z+L)^{2}]^\frac{3}{2}}\Bigg\}\\ &+\frac{\hat{\mathbf{z}}\varepsilon_{d}}{(x^2+\varepsilon_{d} y^2)}\Bigg\{\frac{(z-L)^{2}}{[x^2+\varepsilon_{d} y^2+\varepsilon_{d}(z-L)^{2}]^\frac{3}{2}}-\frac{(z+L)^{2}}{[x^2+\varepsilon_{d} y^2+\varepsilon_{d}(z+L)^{2}]^\frac{3}{2}}\Bigg\}\\ &+\frac{\hat{\mathbf{z}}}{3(x^2+\varepsilon_{d} y^2)}\Bigg\{\frac{(z-L)}{[x^2+\varepsilon_{d} y^2+\varepsilon_{d}(z-L)^{2}]^\frac{1}{2}}-\frac{(z+L)}{[x^2+\varepsilon_{d} y^2+\varepsilon_{d}(z+L)^{2}]^\frac{1}{2}}\Bigg\}\Bigg]\,. \end{aligned}$$
Let us recall that $\mathbf {H}(\mathbf {r})\approx 0$ in the near zone.5.2 Far field except at optic axis
In the far field, the expression of electric field can be found by substituting Eqs. (21) and (51) into Eq. (7), and its simplified expression is given by
(56)$$\begin{aligned} \mathbf{E}(\mathbf{r})&=\frac{\omega^{2}\mu_{o}\mu_{b}{p}_{o}}{4\pi}\Bigg\{\varepsilon_{d}\int_{-L}^{L}\frac{\Big[(x^{2}y\mathbf{\hat{y}}-xy^{2}\mathbf{\hat{x}})(z-z')+x^{2}\mathbf{\hat{z}}(z-z')^{2}-x\mathbf{\hat{x}}(z-z')^{3}\Big]}{y^{2}+(z-z')^{2}}\\ &\times\frac{\exp({i}k_{o}n_{o}R_{ez})}{{R}_{ez}^3} dz^{\prime}+\int_{-L}^{L}\Bigg(\frac{y^{2}\hat{\mathbf{z}}-y(z-z^{\prime})\hat{\mathbf{y}}}{y^{2}+(z-z^{\prime})^{2}}\Bigg)\frac{\exp({i}k_{o}n_{o}R_{z})}{R_{z}} dz^{\prime}\Bigg\}\,. \end{aligned}$$
To solve these integrals we use far field approximation. Using Eq. (16) with $\hat {\mathbf {c}}=\hat {\mathbf {{x}}}$ , we get (57)$$R_{ez}=\Big[r_{e}^{2}+\varepsilon_{d}z^{\prime}(z^{\prime}-2z)\Big]^\frac{1}{2},$$
where $r_{e}$ is given by Eq. (35).In the far-field, our point of observation is far away from the dipole, i.e., $r_{e}\gg z^{\prime }$, so in this limit, we can neglect the higher order terms in $z'/r_e$ in the binomial expansion of (57) and have
(58)$$R_{ez}\simeq r_{e}-\varepsilon_{d}\frac{zz'}{r_{e}}\,$$
in the exponential term, whereas we can approximate $R_{ez}\simeq r_{e}\,$ in the denominator. Similarly, (59)$$R_{z}=\Big[r^2+{z^{\prime}(z^{\prime}-2z)}\Big]^\frac{1}{2}\,$$
can be approximated as (60)$$R_{z}\simeq r-\frac{zz'}{r}$$
in the exponential term and $R_{z}\simeq r$ in the denominator. Furthermore, (61)$$y^2+(z-z')^2\sim y^2+z^2$$
since $y^2+z^2>>L$ in the far field at a point other than the optic axis. Therefore, Eq. (56) becomes (62)$$\begin{aligned} \mathbf{E}(\mathbf{r})&\simeq\frac{\omega^{2}\mu_{o}\mu_{b}{p}_{o}}{4\pi}\Bigg\{\frac{\varepsilon_{d}\exp(i k_{o} n_{o} r_{e})}{r_{e}^{3}(y^2+z^2)}\Big[(x^2y\mathbf{\hat{y}}-xy^2\mathbf{\hat{x}})I_{4}+x^2\mathbf{\hat{z}}I_{5}-x\mathbf{\hat{x}}I_{6}\Big]\\ &+\frac{\exp(i k_{o} n_{o} r)}{r(y^2+z^2)}(y^2\mathbf{\hat{z}}I_{7}-y\mathbf{\hat{y}}I_{8})\Bigg\}\,, \end{aligned}$$
where the integrals $I_{4}$ to $I_{8}$ are derived in the appendix.After substituting the solutions of integrals $I_{4}$ to $I_{8}$ in Eq. (62) and using Eq. (37), the final expression for the electric field is
(63)$$\mathbf{E}=\mathbf{E}_{o}+\mathbf{E}_{e},$$
where (64)$$\mathbf{E}_{o}(\mathbf{r})=\frac{k_{o}\mu_{b}p_{o}}{2\pi\varepsilon_{o}n_{o}r}\Big(-\hat{\mathbf{y}}+\tan\theta\sin\phi\hat{\mathbf{z}}\Big)\frac{\exp(ik_{o}n_{o}r)\sin\theta\sin\phi}{(\sin^2\theta\sin^2\phi+\cos^2\theta)}\sin( k_{o}n_{o}L\cos\theta)\,$$
represents the ordinary wave and (65)$$\begin{aligned} \mathbf{E}_{e}(\mathbf{r})&=\frac{k_{o}\mu_{b}p_{o}}{2\pi\varepsilon_{o}n_{o}r}\Big[-(\sin^2\theta\sin^2\phi+\cos^2\theta)\hat{\mathbf{x}}+\sin^2\theta\cos\phi\sin\phi\hat{\mathbf{y}}\\ &+\sin\theta\cos\theta\cos\phi\hat{\mathbf{z}}\Big]\frac{\exp(ik_{o}n_{o}r\Phi)\sin\theta\cos\phi}{\Phi^2(\sin^2\theta\sin^2\phi+\cos^2\theta)}\sin\Big(\frac{\varepsilon_{d} k_{o}n_{o}L\cos\theta}{\Phi}\Big)\, \end{aligned}$$
represents the extraordinary wave.Now the expression for the magnetic field in the far field with $\hat {\mathbf {c}}=\hat {\mathbf {{x}}}$ can be found by substituting Eqs. (22) and (51) into Eq. (8) as
(66)$$\begin{aligned} \mathbf{H}(\mathbf{r})&={k}_{o}{n}_{o}\omega{p}_{o}\int_{-L}^{L}\Bigg\{{g}_{o}(\mathbf{R}_{z})\frac{[\mathbf{R}_{z}\times(\mathbf{R}_{z}\times\hat{\mathbf{x}})](\mathbf{R}_{z}\times\hat{\mathbf{x}})\cdot\hat{\mathbf{z}}}{{R}_{z}(\mathbf{R}_{z}\times\hat{\mathbf{x}})\cdot(\mathbf{R}_{z}\times\hat{\mathbf{x}})}\\ &-\varepsilon_{d}{g}_{e}(\mathbf{R}_{z})\frac{(\mathbf{R}_{z}\times\hat{\mathbf{x}})[\mathbf{R}_{z}\times(\mathbf{R}_{z}\times\hat{\mathbf{x}})]\cdot\hat{\mathbf{z}}}{{R}_{ez}(\mathbf{R}_{z}\times\hat{\mathbf{x}})\cdot(\mathbf{R}_{z}\times\hat{\mathbf{x}})}\Bigg\}dz^{\prime}\,. \end{aligned}$$
Using Eq. (15) in Eq. (66), we get (67)$$\begin{aligned} \mathbf{H}(\mathbf{r})&=\frac{{k}_{o}{n}_{o}\omega{p}_{o}}{4\pi}\Bigg\{\int_{-L}^{L}\frac{\exp({i}k_{o}n_{o}R_{z})}{{R}_{z}^{2}}\frac{[-xy^{2}\hat{\mathbf{y}}+y^{3}\hat{\mathbf{x}}-xy(z-z^{\prime})\hat{\mathbf{z}}+y(z-z^{\prime})^{2}\hat{\mathbf{x}}]}{y^{2}+(z-z^{\prime})^{2}}dz^{\prime} \\ &-\varepsilon_{d}\int_{-L}^{L}\frac{\exp({i}k_{o}n_{o}R_{ez})}{R_{ez}^{2}}\frac{[-xy(z-z^{\prime})\hat{\mathbf{z}}+x(z-z^{\prime})^{2}\hat{\mathbf{y}}]}{y^{2}+(z-z^{\prime})^{2}}dz^{\prime}\Bigg\}\,. \end{aligned}$$
Just like the electric field, the magnetic field in the far field can be approximated as (68)$$\begin{aligned} \mathbf{H}(\mathbf{r})&\simeq\frac{{k}_{o}{n}_{o}\omega{p}_{o}}{4\pi}\Bigg\{\frac{\exp({i}k_{o}n_{o}r)}{{r}^{2}(y^2+z^2)}\Big[(y^{3}\hat{\mathbf{x}}-xy^{2}\hat{\mathbf{y}})I_{7}-xy\hat{\mathbf{z}}I_{8}+y\hat{\mathbf{x}}I_{9}\Big]\\ &-\frac{\varepsilon_{d}\exp({i}k_{o}n_{o}r_{e})}{r_{e}^{2}(y^2+z^2)}(-xy\hat{\mathbf{z}}I_{4}+x\hat{\mathbf{y}}I_{5})\Bigg\}\,, \end{aligned}$$
where the integral $I_{9}$ in the far zone is derived in the appendix.Substituting the values of $I_{4}$, $I_{7}$, $I_{8}$, and $I_{9}$ in Eq. (68) along with the use of Eqs. (37) and (38), the final expression for the magnetic field is given by
(69)$$\mathbf{H}=\mathbf{H}_{o}+\mathbf{H}_{e},$$
where $\mathbf {H}_{o}$ represents the ordinary wave given by (70)$$\begin{aligned} \mathbf{H}_{o}(\mathbf{r})&=\frac{k_{o}p_{o}c}{2\pi r}\Big[(\sin^2\theta\sin^2\phi+\cos^2\theta)\hat{\mathbf{x}}-\sin^2\theta\cos\phi\sin\phi\hat{\mathbf{y}}\\ &-\sin\theta\cos\theta\cos\phi\hat{\mathbf{z}}\Big]\frac{\exp(ik_{o}n_{o}r)\sin\theta\sin\phi}{\cos\theta(\sin^2\theta\sin^2\phi+\cos^2\theta)}\sin(k_{o}n_{o}L\cos\theta)\,, \end{aligned}$$
and $\mathbf {H}_{e}$ represents the extraordinary wave given by (71)$$\begin{aligned} \mathbf{H}_{e}(\mathbf{r})&=\frac{k_{o}p_{o}c}{2\pi r}\Big(-\cos\theta\hat{\mathbf{y}}+\sin\theta\sin\phi\hat{\mathbf{z}}\Big)\frac{\exp(ik_{o}n_{o}r\Phi)\sin\theta\cos\phi}{\Phi(\sin^2\theta\sin^2\phi+\cos^2\theta)}\sin\Big(\frac{\varepsilon_{d} k_{o}n_{o}L\cos\theta}{\Phi}\Big)\,.\\ & \end{aligned}$$
Since the electric field and magnetic field satisfy the orthogonality relations [26] (72)$$\hat{\mathbf{r}}\cdot(\mathbf{E}_{e}\times\mathbf{H}_{o}^{*})=0,\quad \hat{\mathbf{r}}\cdot(\mathbf{E}_{o}\times\mathbf{H}_{e}^{*})=0,$$
we can find the total radiated power by adding the radiated power of ordinary and extraordinary wave separately, i.e., (73)$$\frac{dP}{d\Omega}=\frac{dP_{o}}{d\Omega}+\frac{dP_{e}}{d\Omega},$$
where (74)$$\frac{dP_{o}}{d\Omega}=\frac{1}{2}\mathbf{\hat{r}}\cdot\rm{Re}(\mathbf{E}_{o}\times\mathbf{H_{o}^{*}})r^{2}\,$$
and (75)$$\frac{dP_{e}}{d\Omega}=\frac{1}{2}\mathbf{\hat{r}}\cdot\rm{Re}(\mathbf{E}_{e}\times\mathbf{H_{e}^{*}})r^{2}\,.$$
Substituting the expression for $\mathbf {E}_{o}$ and $\mathbf {H}_{o}$ from Eqs. (64) and (70) into Eq. (74) and converting into spherical coordinates, we get (76)$$\frac{dP_{o}}{d\Omega}=\frac{k_{o}^{2}\mu_{b}p_{o}^{2}c}{8\pi^{2}\varepsilon_{o}n_{o}}\frac{\sin^2\theta\sin^2\phi}{\cos^2\theta(\sin^2\theta\sin^2\phi+\cos^2\theta)}\sin^2(k_{o}n_{o}L\cos\theta)\,.$$
Similarly, by substituting the expression for $\mathbf {E}_{e}$ and $\mathbf {H}_{e}$ from Eqs. (65) and (71) into Eq. (75) and converting into spherical coordinates, we get (77)$$\frac{dP_{e}}{d\Omega}=\frac{k_{o}^{2}\mu_{b}p_{o}^{2}c}{8\pi^{2}\varepsilon_{o}n_{o}}\frac{\sin^2\theta\cos^2\phi}{\Phi^3(\sin^2\theta\sin^2\phi+\cos^2\theta)}\sin^2\Big(\frac{\varepsilon_{d} k_{o}L n_{o}\cos\theta}{\Phi}\Big)\,.$$
The total time-averaged power radiated per unit solid angle by the dipole is given by substituting Eqs. (76) and (77) into Eq. (73) as (78)$$\begin{aligned} \frac{dP}{d\Omega}&=\frac{k_{o}^{2}\mu_{b}p_{o}^{2}c}{8\pi^{2}\varepsilon_{o}n_{o}}\frac{\sin^2\theta}{(\sin^2\theta\sin^2\phi+\cos^2\theta)}\Bigg[\frac{\cos^2\phi}{\Phi^{3}}\sin^2\Big(\frac{\varepsilon_{d} k_{o}L n_{o}\cos\theta}{\Phi}\Big)\\ &+\frac{\sin^2\phi}{\cos^2\theta}\sin^2(k_{o}n_{o}L\cos\theta)\Bigg]. \end{aligned}$$
When we substitute $\varepsilon _{d}=1$ (giving $\Phi =1$), Eqs. (50) and (78) reduce to the exactly same results.5.3 Far field at optic axis
At the optic axis ($x$ axis), $\#R=R\hat {\mathbf {x}}=R\hat {\mathbf {c}}$ ($\theta =\pi /2$, $\phi =0,\pi$) and the fields and the power has to be computed carefully using the limiting procedure [23]. The dyadic Green functions at the optic axis in the far zone are given by [23]
(79)$$\underline{\underline{G}}^{ee}(\mathbf{R})=i\omega\mu_{o}\mu_{b}\Bigg[\Big(\varepsilon_{a}\underline{\underline{\varepsilon}}_{r}^{-1}-\mathbf{\hat{R}}\mathbf{\hat{R}}\Big)+\frac{\varepsilon_{b}-\varepsilon_{a}}{2\varepsilon_{b}}\Big(\underline{\underline{I}}-\hat{\mathbf{c}}\hat{\mathbf{c}}\Big)\Bigg]g_{o}(\mathbf{R})\,$$
and (80)$$\underline{\underline{G}}^{me}(\mathbf{R})=\frac{ik_{o}n_{o}(\varepsilon_{d}+1)}{2}g_{o}(\mathbf{R})\hat{\mathbf{c}}\times\underline{\underline{I}}\,.$$
The expression for electric field can be found by substituting Eqs. (51) and (79) with $\hat {\mathbf {R}}=\hat {\mathbf {x}}$ in Eq. (7), and is given by (81)$$\mathbf{E}(\mathbf{r})=\frac{k_{o}^2\mu_{b}p_{o}L}{4\pi\varepsilon_{o}}(\varepsilon_{d}+1)\frac{\exp(ik_{o}n_{o}r)}{r}\mathbf{\hat{z}}\,.$$
Similarly, the expression for the magnetic field can be found by substituting Eqs. (51) and (80) in Eq. (8), and is given by (82)$$\mathbf{H}(\mathbf{r})=-\frac{k_{o}^2n_{o}p_{o}cL}{4\pi}(\varepsilon_{d}+1)\frac{\exp(ik_{o}n_{o}r)}{r}\mathbf{\hat{y}}\,.$$
Now by substituting Eqs. (81) and (82) into Eq. (45), the expression of the time-averaged power radiated per unit solid angle by the dipole is given by (83)$$\frac{dP}{d\Omega}=\frac{k_{o}^4\mu_{b}n_{o}p_{o}^2cL^{2}}{32\pi^2\varepsilon_{o}}(\varepsilon_{d}+1)^2\,.$$
When we substitute $\varepsilon _{d}=1$, Eqs. (50) with $\phi =0$ and $\theta =\pi /2$ give the same result as of Eq. (83) for the isotropic medium.6. Numerical results and discussion
The analytical results obtained for the far-field radiation patterns given by Eqs. (46), (50), (76)–(78), and (83) constitute the main results and their elegant closed forms make them usable as it is. To illustrate their use, we present representative numerical results for radiations in rutile with $\varepsilon _a=8.427$, $\varepsilon _b=6.843$, $\mu _b=1$ at $\lambda _{o}=0.584\,\mu$m [26] for a dipole with $p_0=1/\omega$ and length $L=0.1\lambda _o$ and $0.2\lambda _o$.
The far-field radiation patterns of extraordinary waves emitted by a Hertzian dipole aligned with the optic axis ($z$ axis) are shown in Fig. 4. The dipole is placed in uniaxial medium (rutile) when $L=0.1 \lambda _{o}$ and $L=0.2 \lambda _{o}$. The plot is given only for $0\leq \theta \leq \pi$ and $\pi /2\leq \phi \leq 3\pi /2$ for easier visulaization since the pattern is independent of $\phi$, as can be seen from Eq. (50). The figure shows that the pattern is like that of a dipole in an isotropic medium and its directivity increases as the length of the dipole increases.
When the dipole is perpendicular to the optic axis, the radiation pattern of ordinary waves are shown in Fig. 5 for the same medium with $L=0.1 \lambda _{o}$ and $L=0.2 \lambda _{o}$. Again the plot is provided only for $0\leq \theta \leq \pi$ and $\pi /2\leq \phi \leq 3\pi /2$ since the pattern as given by Eq. (76) is symmetric around $xz$ plane as it should be since the dipole and the optic axis define this plane. Its clear from the figure that the radiations in the direction of optic axis are suppressed, though not zero. When we increase the length of dipole, the radiation profile becomes more directive in a direction perpendicular to the dipole.
The radiation pattern of the extraordinary was from the dipole perpendicular to the optic axis are shown in Fig. 6 The figure shows that the emission is highly directive along the optic axis, though the radiations are suppressed along the optic axis in a plane perpendicular to the dipole. Furthermore, there are no radiations emitted along the $y$ axis, a direction perpendicular to both the dipole and the optic axis. When the length of the dipole is increased, the directivity of the radiations increases much more than the previous two cases.
7. Comparison with the point dipole
To compare the results of the Hertzian dipole with those of a point dipole, let us consider a point electric dipole with the electric current density [1]
(84)$$\mathbf{J}_{e}(\mathbf{r'})=-i\omega p_{o}\delta(x')\delta(y')\delta(z')\hat{\mathbf{z}}\,.$$
When both the optic axis and the point dipole are parallel to the $z$ axis, the time-averaged power radiated per unit solid angle can be found as (85)$$\frac{dP}{d\Omega}=\frac{k_{o}^{4}\mu_{b}n_{o}p_{o}^{2}c\varepsilon_{d}^2}{32\pi^{2}\varepsilon_{o}\Theta^5}\sin^2\theta\,$$
in the spherical coordinates. Let us note that $L\rightarrow 0$ in Eq. (50) results in the same form as that of Eq. (85).When the point-dipole is parallel to the $z$ axis and the optic axis parallel to the $x$ axis, the time-averaged power per unit solid angle is given as
(86)$$\frac{dP_{o}}{d\Omega}=\frac{k_{o}^{4}\mu_{b}n_{o} p_{o}^{2}c}{32\pi^{2}\varepsilon_{o}}\frac{\sin^2\theta\sin^2\phi}{\sin^2\theta\sin^2\phi+\cos^2\theta}\,.$$
for the ordinary waves, and (87)$$\frac{dP_{e}}{d\Omega}=\frac{k_{o}^{4}\mu_{b} n_{o} p_{o}^{2}c\varepsilon_{d}^2}{32\pi^{2}\varepsilon_{o}\Phi^5}\frac{\sin^2\theta\cos^2\phi\cos^2\theta}{\sin^2\theta\sin^2\phi+\cos^2\theta}\,.$$
for the extraordinary waves. The total time-averaged power radiated per unit solid angle by the point dipole is given as (88)$$\frac{dP}{d\Omega}=\frac{k_{o}^{4}\mu_{b}n_{o}p_{o}^{2}c}{32\pi^{2}\varepsilon_{o}}\left(\frac{\varepsilon_{d}^2\cos^2\theta\cos^2\phi}{\Phi^5}+\sin^2\phi\right)\frac{\sin^2\theta}{\sin^2\theta\sin^2\phi+\cos^2\theta}.{}$$
The Hertzian-dipole results of Eq. (78) reduce to the same form as the point dipole of Eq. (88) when $L\rightarrow 0$ in Eq. (78).For comparison of the radiation patterns of the point dipole and the Hertzian dipole, the far-field radiation pattern of the point dipole are presented in Fig. 7 and 8 when the dipole is, respectively, parallel and perpendicular to the optics axis, for the same parameters as those for the Hertzian dipole. A comparison Fig. 7 with Fig. 4 shows that the both pattern looks similar, but the directivity is enhanced by the increase in length $2L$ of the Hertzian dipole. The comparison of Fig. 8 with Figs. 5 and 6 draws the same conclusion that the Hertzian dipole allows the control of directivity of the radiation by choosing the length of the dipole.
8. Concluding remarks
The near-field electric field and the far-field radiation pattern of a Hertzian dipole in uniaxial dielectric medium were analytically derived and the far-field numerical results were presented for a chosen uniaxial material. The dipole was taken to be along and perpendicular to the optic axis. When the dipole was parallel to the optic axis, only extraordinary waves were emitted in the far field. When the dipole was perpendicular to the optic axis, both ordinary and extraordinary waves were emitted but the radiations along the optic axis were suppressed for both the ordinary and extraordinary waves. For the latter case, there was no emission of extraordinary waves perpendicular to the dipole and the optic axis in the far field. For all the cases, the directivity of the radiation pattern increased significantly with the increase in the length of the dipole. A comparison with the results of the point dipole showed that the the length of the Hertzian dipole plays a significant role in the directivity of the radiation pattern.
Appendix
In this appendix, we present the derivations of the integrals used in the formulation of the far-field radiations. In general, we solved the integrals in the Cartesian coordinates, converted them into the spherical coordinates, and then retained only the leading term.
Let us begin with
(89)$$I_{1}=\int_{-L}^{L}\exp(-i s x x')dx^\prime=\frac{2\sin(sLx)}{sx},$$
where
(90)$$s=\frac{k_{o}n_{o}}{r_{e}}\,.$$
By substituting the expressions for x and
$r_{e}$ from Eqs. (
37) and (
38) into Eq. (
89), we get
(91)$$I_{1}=\frac{2\Phi}{k_{o}n_{o}\sin\theta\cos\phi}\sin\left(\frac{k_{o}n_{o}L\sin\theta\cos\phi}{\Phi}\right)\,.$$
The second integral is
(92)$$\begin{aligned} I_{2}&=\int_{-L}^{L}(x-x^{\prime})\exp(-is xx')dx^{\prime}\\ &=\frac{-2isLx\cos( Lsx)+2(i+ sx^{2})\sin( Lsx)}{s^{2}x^{2}}\,. \end{aligned}$$
By substituting the expressions for x and
$r_{e}$ from Eqs. (
37) and (
38) into Eq. (
92) and keeping only the dominant term as
$r\rightarrow \infty$, we get
(93)$$I_{2}=\frac{2r\Phi}{k_{o}n_{o}}\sin\Bigg(\frac{k_{o} n_{o}L\sin\theta\cos\phi}{\Phi}\Bigg)\,.$$
The third integral is
(94)$$\begin{aligned} I_{3}&=\int_{-L}^{L}(x-x')^2\exp(-is xx')dx^{\prime}\\ &=\frac{i}{s^{3}x^{3}}\Big\{\exp(-isLx)[-2-2is(L-x)x+s^{2}(L-x)^{2}x^{2})]\\ &-\exp(isLx)[-2+2is(L+x)x+s^{2}(L+x)^{2}x^{2}]\Big\}\,. \end{aligned}$$
By converting to spherical coordinates and retaining only the dominant term as
$r\rightarrow \infty$, we get
(95)$$I_{3}=\frac{2r^{2}\Phi\sin\theta\cos\phi}{k_{o}n_{o}}\sin\Bigg(\frac{k_{o} n_{o}L\sin\theta\cos\phi}{\Phi}\Bigg)\,.$$
The fourth integral is
(96)$$\begin{aligned} I_{4}&=\int_{-L}^{L}(z-z^{\prime})\exp(-is\varepsilon_{d} zz')dz^{\prime}\\ &=\frac{-2i\varepsilon_{d} sLz\cos(\varepsilon_{d} Lsz)+2(i+\varepsilon_{d} sz^{2})\sin(\varepsilon_{d} Lsz)}{\varepsilon_{d}^{2}s^{2}z^{2}}\,. \end{aligned}$$
By substituting the expressions for z and
$r_{e}$ from Eqs. (
37) and (
38) into Eq. (
96) and keeping only the dominant term as
$r\rightarrow \infty$, we get
(97)$$I_{4}=\frac{2r\Phi}{\varepsilon_{d} k_{o}n_{o}}\sin\Bigg(\frac{\varepsilon_{d} k_{o} n_{o}L\cos\theta}{\Phi}\Bigg)\,.$$
The fifth integral is
(98)$$\begin{aligned} I_{5}&=\int_{-L}^{L}(z-z')^2\exp(-is\varepsilon_{d} zz')dz^{\prime}\\ &=\frac{i}{s^{3}z^{3}\varepsilon_{d}^{3}}\Big\{\exp(-isLz\varepsilon_{d})[-2-2is(L-z)z\varepsilon_{d}+s^{2}(L-z)^{2}z^{2}\varepsilon_{d}^{2}]\\ &-\exp(isLz\varepsilon_{d})[-2+2is(L+z)z\varepsilon_{d}+s^{2}(L+z)^{2}z^{2}\varepsilon_{d}^{2}]\Big\}\,. \end{aligned}$$
By converting to spherical coordinates and retaining only the dominant term as
$r\rightarrow \infty$, we get
(99)$$I_{5}=\frac{2r^{2}\Phi\cos\theta}{\varepsilon_{d} k_{o}n_{o}}\sin\Bigg(\frac{\varepsilon_{d} k_{o} n_{o}L\cos\theta}{\Phi}\Bigg)\,.$$
The sixth integral is
(100)$$\begin{aligned} I_{6}&=\int_{-L}^{L}(z-z')^3\exp(-is\varepsilon_{d} zz')dz^{\prime}\\ &=\frac{1}{s^{4}z^{4}\varepsilon_{d}^{4}}\Big\{\exp(-isLz\varepsilon_{d})\Big[6+i6s(L-z)z\varepsilon_{d}-3s^2(L-z)^2 z^2\varepsilon_{d}^2-i s^{3}(L-z)^{3}z^{3}\varepsilon_{d}^{3}\Big]\\ &-\exp(isLz\varepsilon_{d})\Big[6-i6s(L+z)z\varepsilon_{d}-3s^2(L+z)^2 z^2\varepsilon_{d}^2+i s^{3}(L+z)^{3}z^{3}\varepsilon_{d}^{3}\Big]\Big\}\,. \end{aligned}$$
By converting to spherical coordinates and retaining only the dominant term as
$r\rightarrow \infty$, we get
(101)$$I_{6}=\frac{2r^{3}\Phi\cos^{2}\theta}{\varepsilon_{d} k_{o}n_{o}}\sin\Bigg(\frac{\varepsilon_{d} k_{o} n_{o}L\cos\theta}{\Phi}\Bigg)\,.$$
The seventh integral
(102)$$I_{7}=\int_{-L}^{L}\exp(-i\tau zz')dz^{\prime}\,$$
is similar to
$I_1$ with
(103)$$\tau=\frac{k_{o}n_{o}}{r}\,.$$
Performing integration and keeping only the dominant term as
$r\rightarrow \infty$, the final expression for
$I_{7}$ in spherical coordinates is
(104)$$I_{7}=\frac{2\sin(k_{o}n_{o}L\cos\theta)}{k_{o}n_{o}\cos\theta}\,.$$
The eighth integral is similar to
$I_4$, so
(105)$$I_{8}=\int_{-L}^{L}(z-z')\exp(-i\tau zz')dz^{\prime}\,=\frac{2r\sin(k_{o}n_{o}L\cos\theta)}{k_{o}n_{o}}\,.$$
The integral
$I_9$ is similar to
$I_5$ so
(106)$$I_{9}=\int_{-L}^{L}(z-z')^{2}\exp(-i\tau zz')dz^{\prime}=\frac{2r^{2}\cos\theta\sin(k_{o}n_{o}L\cos\theta)}{k_{o}n_{o}}\,.{}$$
Funding
Higher Education Commision, Pakistan (HEC) (NRPU 2016–5905).
References
1. H. C. Chen, Theory of Electromagnetic Waves: A Coordinate-Free Approach (McGraw Hill, 1983), Ch. 10.
2. H. C. Chen, “Dyadic green’s function and radiation in a uniaxially anisotropic medium,” Int. J. Electron. 35(5), 633–640 (1973). [CrossRef]
3. L. B. Felsen and N. Marcuvitz, Radiation and Scattering of Waves (IEEE, 1994).
4. K. Korzeb, M. Gajc, and D. A. Pawlak, “Compendium of natural hyperbolic materials,” Opt. Express 23(20), 25406–25424 (2015). [CrossRef]
5. M. Born and E. Wolf, Principles of Optics: Electromagnetic Theory of Propagation, Interference and Diffraction of Light (Pergamon, 1970).
6. K. G. Balmain, A. A. E. Luttgen, and P. C. Kremer, “Resonance cone formation, reflection, refraction, and focusing in a planar anisotropic metamaterial,” IEEE Antennas Wirel. Propag. Lett. 1, 146–149 (2002). [CrossRef]
7. T. Taubner, D. Korobkin, Y. Urzhumov, G. Shvets, and R. Hillenbrand, “Near-field microscopy through a sic superlens,” Science 313(5793), 1595 (2006). [CrossRef]
8. J. Yao, X. Yang, X. Yin, G. Bartal, and X. Zhang, “Three-dimensional nanometer-scale optical cavities of indefinite medium,” Proc. Natl. Acad. Sci. USA 108(28), 11327–11331 (2011). [CrossRef]
9. Y. Guo, C. L. Cortes, S. Molesky, and Z. Jacob, “Broadband super-planckian thermal emission from hyperbolic metamaterials,” Appl. Phys. Lett. 101(13), 131106 (2012). [CrossRef]
10. U. Tumkur, L. Gu, J. K. Kitur, E. E. Narimanov, and M. A. Noginov, “Control of absorption with hyperbolic metamaterials,” Appl. Phys. Lett. 100(16), 161103 (2012). [CrossRef]
11. Z. Liu, H. Lee, Y. Xiong, C. Sun, and X. Zhang, “Far-field optical hyperlens magnifying sub-diffraction-limited objects,” Science 315(5819), 1686 (2007). [CrossRef]
12. H. Kuzmany, Solid State Spectroscopy (Springer-Verlag, 2009).
13. A. Eroglu and J. Kyoon, “Far field radiation from an arbitrarily oriented hertzian dipole in the presence of a layered anisotropic medium,” IEEE Trans. Antennas Propag. 53(12), 3963–3973 (2005). [CrossRef]
14. A. Lakhtakia, V. K. Varadan, and V. V. Varadan, “Radiation and canonical sources in uniaxial dielectric media,” Int. J. Electron. 65(6), 1171–1175 (1988). [CrossRef]
15. L. Alexeyeva, I. Kanymgaziyeva, and S. Sautbekov, “Generalized solutions of maxwell equations for crystals with electric and magnetic anisotropy,” J. Electromagn. Waves Appl. 28(16), 1974–1984 (2014). [CrossRef]
16. L. Tsang, E. Njoku, and J. A. Kong, “Microwave thermal emission from a stratified medium with nonuniform temperature distribution,” J. Appl. Phys. 46(12), 5127–5133 (1975). [CrossRef]
17. J. A. Kong, “Electromagnetic fields due to dipole antennas over stratified anisotropic media,” Geophysics 37(6), 985–996 (1972). [CrossRef]
18. Y. S. Kwon and J. J. H. Wang, “Computation of hertzian dipole radiation in stratified uniaxial anisotropic media,” Radio Sci. 21(6), 891–902 (1986). [CrossRef]
19. C. M. Tang, “Electromagnetic fields due to dipole antennas embedded in stratified anisotropic media,” IEEE Trans. Antennas Propag. 27(5), 665–670 (1979). [CrossRef]
20. S. Ali and S. Mahmoud, “Electromagnetic fields of buried sources in stratified anisotropic media,” IEEE Trans. Antennas Propag. 27(5), 671–678 (1979). [CrossRef]
21. N. Bellyustin and V. Dokuchaev, “Generation of electromagnetic waves by distributed currents in an anisotoipic medium,” Radiophys. Quantum Electron. 18(1), 10–17 (1975). [CrossRef]
22. T. N. C. Wang and T.-L. Wang, “Radiation resistance of small electric and magnetic antennas in a cold uniaxial plasma,” IEEE Trans. Antennas Propag. 20(6), 796–798 (1972). [CrossRef]
23. M. Faryad and A. Lakhtakia, Infinite-Space Dyadic Green Functions in Electromagnetism (Morgan & Claypool, 2018).
24. J. D. Jackson, Classical Electrodynamics, 3rd ed. (Wiley, 1999).
25. A. Zangwill, Modern Electrodynamics (Cambridge, 2012), Ch. 20.
26. A. Yariv and P. Yeh, Optical Waves in Crystals: Propagation and Control of Laser Radiation (John Wiley & Sons, 1984).
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